March 5, 2002
(2) A object moving on a straight line has a (nonconstant) acceleration of 6t feet per second per second at time t. It travels a distance of 33 feet between times t = 0 and t = 3. Find the object's velocity function v(t).
a(t) = 6t
v(t) = = 3+C (We just need to find C)
s(t) = = +Ct +K
From s(0) = 0 we get K = 0, so s(t) = +Ct Now, 33 = s(3) = gives the equation 33 = 27 + 3C, or 3C = 6, so C = 2.
Thus v(t) = 3+2
(4) Find the average value of the function on the interval [0, ln(2)].
Using the formula for average value, the average value is
Over the interval [6,2], the graph of y = |x| forms two triangles. Adding their areas, we get the answer of
(1/2)(2)(2) + (1/2)(6)(6) = 2 + 18 = 20.
(6) The graph of a function f(x) has slope at each point (x, f(x)).
Also, the graph of f(x) passes through the point (1, 5). Find the function f(x).
The information says slope = f '(x) = .
This means f(x) =
Thus f(x) =
To determine f completely, we just need to find C. To do this, use the fact that f(1)=5.
5= 0 + 3/4 +C
Therefore C = 17/4, and the answer to the question is f(x) =
(7) The expression represents a definite integral over the interval [3, 5]. Write the definite integral. (You do not need to find its value.) ANSWER: